The equation of tangent to the curve  $y=x^4-6x^3+13x^2-10x+5$   at the point (0,5) is

$$\begin{gathered} y=x^4-6x^3+13x^2-10x+5 at (0,5) \\ \frac{dy}{dx}=4x^3-6\left(3x^2\right)+3\left(2x\right)-10 \frac{dy}{dx}=4x^3-18x^2+6x-10 m=-10 \end{gathered}$$

Tangent (0,5)                                Normal
$y-5=-10(x-0)$                 $y-5=\frac{-1}{-10}\left(x\right)$

$y-5=-10x$                         $10y-50=x$

$10x+y-5=0$                       $x-10y+50=0$